$\|x\| \leq \|x(h)\|$ for all $x\in A$ implies $\dim A < \infty$.

Question

Let $A\subseteq B(H)$ be a unital inclusion of $C^*$-algebras. Suppose that there is a vector $h \in H$ such that $\|x \| \leq \|x(h)\|$ for all $x \in A$. Does it follow that $A$ is finite-dimensional?

Attempt/observations:

Not really sure how to start. Somehow we want to show that a set of linearly independent vectors in $A$ must be finite.

• Taking $x=1$, we see that $\|h\| \geq 1$.
• Maybe we can show that if $A$ has infinite dimension, then there is $x \in A$ with $\|x\|> \|x(h)\|$, or equivalently $\|x^*x\| > \langle x^*xh,h\rangle.$

Answer 1

The condition forces $A$ to be finite-dimensional.

This is because the inequality will also hold for the double commutant $A''$. Indeed, $A''$ is the sot-closure of $A$. If $x=\lim_{sot}x_j$ and $\|x_j\|\leq\|x_jh\|$, then $$ \|x\|\leq\limsup_j\|x_j\|\leq\limsup_j\|x_jh\|=\|xh\|. $$ Being a von Neumann algebra, $A''$ is the norm closure of the span of its projections. Let $p_1,\ldots,p_m\in A''$ be pairwise orthogonal projections. Then $$ \|h\|^2=\sum_j\|p_kh\|^2\geq m $$ So $A''$ admits only finitely many pairwise orthogonal projections. This forces $A''$ (and, a fortiori, $A$, since they are now equal) to be finite-dimensional.